HDU 4303 树形DP

Hourai Jeweled

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 163840/163840 K (Java/Others)
Total Submission(s): 1149    Accepted Submission(s): 457

Problem Description

Kaguya Houraisan was once a princess of the Lunarians, a race of people living on the Moon. She was exiled to Earth over a thousand years ago for the crime of using the forbidden Hourai Elixir to make herself immortal. Tales of her unearthly beauty led men from all across the land to seek her hand in marriage, but none could successfully complete her trial of the Five Impossible Requests. 

One of these requests is to reckon the value of "Hourai Jeweled (蓬莱の玉の枝)". The only one real treasure Kaguya has, in her possession. As showed in the picture, Hourai Jeweled is a tree-shaped twig. In which, each node is ornamented with a valuable diamond and each edge is painted with a briliant color (only bright man can distinguish the difference). Due to lunarians' eccentric taste, the value of this treasure is calculated as all the gorgeous roads' value it has. The road between two different nodes is said to be gorgeous, if and only if all the adjacent edges in this road has diffenrent color. And the value of this road is the sum of all the nodes' through the road.
Given the value of each node and the color of each edge. Could you tell Kaguya the value of her Hourai Jeweled?

 

 

Input

The input consists of several test cases. 
The first line of each case contains one integer N (1 <= N <= 300000), which is the number of nodes in Hourai Jeweled.
The second line contains N integers, the i-th of which is Vi (1 <= Vi <= 100000), the value of node i.
Each of the next N-1 lines contains three space-separated integer X, Y and Z (1<=X,Y<=N, 1 <= Z <= 100000), which represent that there is an edge between X and Y painted with colour Z. 

 

 

Output

For each test case, output a line containing the value of Hourai Jeweled.

 

 

Sample Input

6

6 2 3 7 1 4

1 2 1

1 3 2

1 4 3

2 5 1

2 6 2

 

 

Sample Output

134

Hint

gorgeous roads are : 1-2 Value: 8 1-3 Value: 9 1-4 Value:13 1-2-6 Value:12 2-1-3 Value:11 2-1-4 Value:15 2-5 Value:3 2-6 Value:6 3-1-4 Value:16 3-1-2-6 Value:15 4-1-2-6 Value:19 5-2-6 Value:7

 

 

Author

BUPT

 

 

Source

2012 Multi-University Training Contest 1

 题意：
n个节点的树，每一个点有权值，每条边有一种颜色，问全部美丽路径的权值之和，美丽路径是相邻的两条边的颜色不一样的路径。

代码：

//作法很好想但实现很难，两条边以上的路径有两种，一种是儿子节点及其后代和父亲链接的路径
//另外一种是兄弟之间链接的路径但处理兄弟之间组成的路径很差处理。dp[i]存i节点以及其后代的
//权值和，cntp[i]存i节点以及其后代中共有多少个节点。好难。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
int n,head[300009],tol,pa[300009],cntp[300009],val[300009],S[300009];
ll dp[300009],vc[300009],vw[300009],ans;
struct node{
    int v,w,next;
}nodes[600009];
void Add(int x,int y,int z){
    nodes[tol].v=y;
    nodes[tol].w=z;
    nodes[tol].next=head[x];
    head[x]=tol++;
}
void dfs(int u,int fa){
    cntp[u]=1;dp[u]=val[u];
    ll sum=0;
    for(int i=head[u];i!=-1;i=nodes[i].next){
        int v=nodes[i].v;
        if(v==fa) continue;
        pa[v]=nodes[i].w;
        dfs(v,u);
    }
    int c=0;
    for(int i=head[u];i!=-1;i=nodes[i].next){
        int v=nodes[i].v;
        if(v==fa) continue;
        cntp[u]+=cntp[v];
        if(!vc[nodes[i].w]) S[++c]=nodes[i].w;
        vc[nodes[i].w]+=cntp[v];
        vw[nodes[i].w]+=dp[v];
        if(nodes[i].w!=pa[u])//更新u
            dp[u]+=dp[v]+1ll*cntp[v]*val[u];
        ans+=dp[v]+1ll*cntp[v]*val[u];//和父亲链接
        sum+=dp[v];
        ans+=((sum-vw[nodes[i].w])*cntp[v]+dp[v]*(cntp[u]-1-vc[nodes[i].w])+1ll*val[u]*cntp[v]*(cntp[u]-1-vc[nodes[i].w]));//兄弟之间链接
    }
    cntp[u]-=vc[pa[u]];//减去颜色冲突的
    while(c){
        vc[S[c]]=0;
        vw[S[c--]]=0;
    }
}
int main()
{
    while(scanf("%d",&n)==1){
        int x,y,z;
        tol=0;
        memset(head,-1,sizeof(head));
        for(int i=1;i<=n;i++) scanf("%d",&val[i]);
        for(int i=1;i<n;i++){
            scanf("%d%d%d",&x,&y,&z);
            Add(x,y,z);
            Add(y,x,z);
        }
        ans=0;pa[1]=0;
        dfs(1,0);
        printf("%I64d\n",ans);
    }
    return 0;
}